Nature Of Electromagnetic Waves - JEE Practice Questions & MCQ

Nature of Electromagnetic Waves

From Maxwell’s equations, we can observe that electric and magnetic fields in an electromagnetic wave are perpendicular to each other, and to the direction of propagation. Also from our discussion of the displacement current, in that capacitor, the electric field inside the plates of the capacitor is directed perpendicular to the plates. The figure given below shows a typical example of a plane electromagnetic wave propagating along the z direction (the fields are shown as a function of the z coordinate, at a given time t). The electric field E_x is along the x-axis, and varies sinusoidally with z, at a given time. The magnetic field B_y is along the y-axis, and again varies sinusoidally with z.The electric and magnetic fields E_x and B_y are perpendicular to each other, and to the direction z of propagation.

Now from the Lorentz equation -

$
\begin{aligned}
& \qquad \vec{F}=q(\vec{E}+\vec{v} \times \vec{B}) \\
& E_z=E z_0 \sin (\omega t-k y) \\
& B_x=B x_0 \sin (\omega t-k y), \text { where } \frac{\omega}{k}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \\
& \text { the frequency and } k=\frac{2 \pi}{\lambda}, \text { where } \lambda \text { is the wavelength. }
\end{aligned}
$

since, $\omega=2 \pi f$, where f is the frequency and

Therefore, $\frac{\omega}{k}=\frac{2 \pi f}{2 \pi / \lambda}=f \lambda$
But $f \lambda$ gives the velocity of the wave. So, $f \lambda=c=\omega k$. So we can write -

$
c=\frac{\omega}{k}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}
$
 

It is also seen from Maxwell's equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as -

$
B_0=\frac{E_o}{c}
$

In a material medium of permittivity $\varepsilon$ and magnetic permeability $\mu$, the velocity of light becomes,

$
v=\frac{1}{\sqrt{\mu \varepsilon}}
$